Expression for Angular Frequency

[zorro]

How is this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R²/4L²)^1/2

(I came across this while solving a problem)

 

[zhermes]

You solve the differential equation for kirchhoffs loop rule.

 

[tiny-tim]

Hi Abdul!

How does this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R²/4L²)^1/2

(I came across this while solving a problem)

Well, without knowing anything about circuits, we can guess that it probably has something to do with the fact that that obviously looks like a determinant,

of Lx² + Rx + 1/C = 0 …​

can you get a differential equation that looks like that, maybe for current?

 

[zorro]

I got the same equation for w in place of x. On solving,

w = (-R +/- (R² -4L/C)^1/2)/2L

I did not get any differential equation.

 

[tiny-tim]

I got the same equation for w in place of x. On solving,

w = (-R +/- (R² -4L/C)^1/2)/2L

Are you sure? that's not the same ω as in your original post …

How is this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R²/4L²)^1/2

I suspect that the R/2L is a separate exponential part

(my differential equation in I was from combining CV_C = Q (so CdV_C/dt = I), V_R = IR, V_L = LdI/dt, and d/dt(V_C + V_R + V_L) = 0)

 

[zorro]

(my differential equation in I was from combining CV_C = Q (so CdV_C/dt = I), V_R = IR, V_L = LdI/dt, and d/dt(V_C + V_R + V_L) = 0)

That will give a second order differential equation, right?
Your expression does not contain any w.

 

[tiny-tim]

It gives I as a function of t …

part of that solution will be harmonic, and ω will be the frequency of the harmonic part.

 

[zorro]

I got
I/C + RdI/dt + Ld²I/dt² = 0

How to solve this?

 

[tiny-tim]

Hi Abdul!

(just got up :zzz: …)

I got
I/C + RdI/dt + Ld²I/dt² = 0

How to solve this?

Don't you know the standard method for these equations? …

write D = d/dt, so the equation becomes

(LD² + RD + 1/C)I = 0,

which you can factor to

(D - R/2L + i√(1/LC + R²/4L²))(D - R/2L - i√(1/LC + R²/4L²))I = 0 …

carry on from there ​

 

[zorro]

Its d²I/dt² not (dI/dt)²

 

[tiny-tim]

That's ok, that is what D² means …

D²(I) = D(D(I)) = d/dt(d/dt(I)) = d/dt(dI/dt) = d²I/dt²

 

[zorro]

I have solved questions in D.E. like -
1 + (dy/dx)² = xdy/dx

by taking dy/dx = p

Do we use p² for both (dy/dx)² and d²y/dx² ?

 

[tiny-tim]

no …

that p equation has an xdy/dx,

the D is for equations with just dy/dx